Giải hệ PT :
\(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}=3\)
\(x+y+z=3\)
Những câu hỏi liên quan
a) Cho x,y,z thỏa mãn x+y+z+xy+yz+zx=6. Tìm Min \(P=x^2+y^2+z^2\)
giải hệ pt : 1) \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}+\sqrt{2-\dfrac{1}{y}}=2\\\dfrac{1}{\sqrt{y}}+\sqrt{2-\dfrac{1}{x}}=2\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^4+x^2y^2+y^4=21\end{matrix}\right.\)
1. Với mọi số thực x;y;z ta có:
\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)
\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)
\(\Rightarrow P\ge3\)
\(P_{min}=3\) khi \(x=y=z=1\)
1.1
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Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)
\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)
\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)
\(\Leftrightarrow a=b\Leftrightarrow x=y\)
Thay vào pt đầu:
\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))
\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)
\(\Rightarrow a=1\Rightarrow x=y=1\)
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2.
\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)
\(\Rightarrow4x^2-10xy+4y^2=0\)
\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu
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1.Giải hệ pt1)hept{begin{cases}frac{1}{x}+frac{1}{y}+frac{1}{z}3xy+yz+zx3frac{1}{1+x+xy}+frac{1}{1+y+yz}+frac{1}{1+z+zx}xend{cases}}2)hept{begin{cases}xy+yz+zx3left(x+yright)left(y+zright)sqrt{3}zleft(1+y^2right)left(y+zright)left(z+xright)sqrt{3}xleft(1+z^2right)end{cases}}3)hept{begin{cases}xy+yz+zx31+x^2left(y+zright)+xyz4y1+y^2left(z+xright)+xyz4zend{cases}}
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1.Giải hệ pt
1)\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\\xy+yz+zx=3\\\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=x\end{cases}}\)
2)\(\hept{\begin{cases}xy+yz+zx=3\\\left(x+y\right)\left(y+z\right)=\sqrt{3}z\left(1+y^2\right)\\\left(y+z\right)\left(z+x\right)=\sqrt{3}x\left(1+z^2\right)\end{cases}}\)
3)\(\hept{\begin{cases}xy+yz+zx=3\\1+x^2\left(y+z\right)+xyz=4y\\1+y^2\left(z+x\right)+xyz=4z\end{cases}}\)
Giải hệ PT sau: \(\hept{\begin{cases}3\sqrt{x}+2\sqrt{y}+\sqrt{z}=\frac{\sqrt{xyz}}{6}\\6\sqrt{xy}+2\sqrt{yz}+3\sqrt{zx}=108+18\sqrt{x+4}+12\sqrt{y+9}+6\sqrt{z+36}\end{cases}}\)
5 .\(\frac{x}{\sqrt{2\left(y^2+z^2\right)-x^2}}=\frac{\sqrt{3}x^2}{\sqrt{3}x\sqrt{2\left(y^2+z^2\right)-x^2}}\ge\frac{\sqrt{3}x^2}{x^2+y^2+z^2}\)
TT=>VT2>=VP2
6.\(1+\sqrt{y-1}\ge1\)
\(\frac{1}{y^2}-\left(x+z\right)^2\le1\)
=>VT1>=VP1
10b pt1\(\Leftrightarrow\left(y-3x\right)\left(y^2-y+1\right)=0\)
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chi. cậu trả lời j vào câu hỏi của tớ vậy???
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mượn 1 tí thôi tại mai bảo là ghi đại vào đâu đấy
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Xem thêm câu trả lời
cho Q= \(\sqrt{x^2-xy+y^2}\)+ \(\sqrt{y^2-yz+z^2}\)+\(\sqrt{z^2-zx+x^2}\) với x,y,z > 0 x+y+z=3
CM : Q ≥ 3
\(x^2-xy+y^2=\dfrac{1}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{x^2-xy+y^2}\ge\sqrt{\dfrac{1}{4}\left(x+y\right)^2}=\dfrac{1}{2}\left(x+y\right)\)
Tương tự: \(\sqrt{y^2-yz+z^2}\ge\dfrac{1}{2}\left(y+z\right)\); \(\sqrt{z^2-zx+x^2}\ge\dfrac{1}{2}\left(z+x\right)\)
Cộng vế:
\(Q\ge\dfrac{1}{2}\left(x+y\right)+\dfrac{1}{2}\left(y+z\right)+\dfrac{1}{2}\left(z+x\right)=x+y+z=3\) (đpcm)
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Cho x,y,z dương. Chứng minh \(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)
\(VT=\sum\sqrt{\frac{1}{2}\left(x^2+2xy+y^2\right)+\frac{1}{2}\left(x^2+y^2\right)}\)
\(VT\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{1}{4}\left(x+y\right)^2}=\sqrt{\frac{3}{4}\left(x+y\right)^2}\)
\(VT\ge\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{\sqrt{3}}{2}\left(y+z\right)+\frac{\sqrt{3}}{2}\left(z+x\right)=\sqrt{3}\left(x+y+z\right)\)
Dấu "=" xảy ra khi \(x=y=z\)
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Cho x, y, z dương. Chứng minh rằng: \(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}.\left(x+y+z\right)\)
Lời giải:
Ta thấy:
\(x^2+xy+y^2=\frac{3}{4}(x^2+2xy+y^2)+\frac{1}{4}(x^2-2xy+y^2)=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\)
\(\geq \frac{3}{4}(x+y)^2\) với mọi $x,y>0$
\(\Rightarrow \sqrt{x^2+xy+y^2}\geq \frac{\sqrt{3}}{2}(x+y)\)
Hoàn toàn tương tự:
\(\sqrt{y^2+yz+z^2}\geq \frac{\sqrt{3}}{2}(y+z); \sqrt{z^2+zx+x^2}\geq \frac{\sqrt{3}}{2}(x+z)\)
Cộng theo vế các BĐT trên và rút gọn:
\(\Rightarrow \sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+xz+x^2}\geq \sqrt{3}(x+y+z)\)
Ta có đpcm.
Dấu "=" xảy ra khi $x=y=z$
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Cho x,y,z dương. Chứng minh \(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)
đặt Afrac{sqrt{yz}}{x+3sqrt{yz}}+frac{sqrt{zx}}{y+3sqrt{zx}}+frac{sqrt{xy}}{z+3sqrt{xy}}Rightarrow1-3Afrac{x}{x+3sqrt{yz}}+frac{y}{y+3sqrt{zx}}+frac{z}{z+3sqrt{xy}}gefrac{x}{x+frac{3}{2}left(y+zright)}+frac{y}{y+frac{3}{2}left(z+xright)}+frac{z}{z+frac{3}{2}left(x+yright)}frac{2x}{2x+3left(y+zright)}+frac{2y}{2y+3left(z+xright)}+frac{2z}{2z+3left(x+yright)}frac{2x^2}{2x^2+3xy+3xz}+frac{2y^2}{2y^2+3yz+3xy}+frac{2z^2}{2z^2+3zx+3yz}gefrac{2left(x+y+zright)^2}{2left(x^2+y^2+z^2right)+6left(xy+yz+zxr...
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đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)
\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)
\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)
\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)
\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)
ta có Pfrac{x^2}{xsqrt{y+3}}+frac{y^2}{ysqrt{z+3}}+frac{z^2}{zsqrt{x+3}}gefrac{left(x+y+zright)^2}{xsqrt{y+3}+ysqrt{z+3}+zsqrt{x+3}}mà left(xsqrt{y+3}+...right)^2leleft(x+y+zright)left(xy+yz+zx+3x+3y+3zright)le3left(9+3right)36 ( vì xy+yz+zx3)xsqrt{y+3}+...le6Rightarrow Pgefrac{9}{6}frac{3}{2}dấu xảy ra xyz1
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ta có P=\(\frac{x^2}{x\sqrt{y+3}}+\frac{y^2}{y\sqrt{z+3}}+\frac{z^2}{z\sqrt{x+3}}\ge\frac{\left(x+y+z\right)^2}{x\sqrt{y+3}+y\sqrt{z+3}+z\sqrt{x+3}}\)
mà \(\left(x\sqrt{y+3}+...\right)^2\le\left(x+y+z\right)\left(xy+yz+zx+3x+3y+3z\right)\le3\left(9+3\right)=36\) ( vì xy+yz+zx<=3)
=>\(x\sqrt{y+3}+...\le6\Rightarrow P\ge\frac{9}{6}=\frac{3}{2}\)
dấu = xảy ra <=> x=y=z=1